The equation x^{2}+ ax + 2023 = 0 has roots x_{1 }= b + 1 and x_{2} = 8b^{2} + 1 for some real number b. Find a + b.

So far I have been trying Vieta's and getting 8b^{3} + 8b^{2} + b +1 = 2023, -a = 8b^{2} + b + 2. And then getting 2023 + a = 8b^3, but then I don't know what to do from here, is my approach totally wrong? Also, can I have the answer with a solution, so I understand how you got the answer?

Guest May 16, 2023

#1**0 **

The sum of the roots of a quadratic equation is given by -b/a, and the product of the roots is given by c/a, where a, b, and c are the coefficients of the quadratic equation.

In this case, we know that the sum of the roots is b + 8b^2 + 2, and the product of the roots is 2023.

We can use these values to find a and b.

(b+1)+(8b^2+1)=b+8b^2+2 2b+2=b+8b^2+2 b=8b^2

We know that b is a real number, so b cannot be 0. Therefore, b = 1/8.

We can now find a.

-b/a+c/a=b+8b^2+2 -1/(8a)+2023/a=b+8b^2+2 -1/(8a)+2023/a=1/8+2023/8 2022/(8a)=2025/8 a=2025/2022

Therefore, a + b =2025/2022+1/8 =2051/2022.

Guest May 16, 2023